I went rummaging through the garbage of the last deluge of bad AI "solutions"
and found this interesting rhombus problem buried there. (incorrectly done by
AI.)

Rhombuses are easier to think about when you draw them diamond-shaped, i.e.,
symmetrical with the horizontal and the vertical. So I will draw the figure
that way instead of the way it's drawn on the site of the given link. 

As you can see from the second figure below, any rhombus can be partitioned into
8 CONGRUENT isosceles triangles (they might look equilateral but that's not
necessarily the case. They are only isosceles, I just accidentally drew them to
look equilateral.)

Anyway, the second figure below shows that ΔTPW is 1/8 of the rhombus 
(area-wise).  That's too obvious to bother wasting time to prove.  

Quadrilateral PWVT is made up of ΔTPW and ΔTVW.  So all that's left is to find
what fraction ΔTVW is of the whole rhombus and add that to 1/8.

     

We will use these two well-known theorems:

Theorem 1: If two triangles are similar, the ratio of their areas is equal to
the square of the ratio of any pair of corresponding sides between them. 

Theorem 2: The intersection of the three medians of a triangle, called the
centroid, is located two-thirds of the distance from each vertex to the midpoint
of the opposite side.  That is to say, the shorter part of each median from the
centroid is 1/2 the longer part from the centroid.  It also says that the
entire median is 3 times the shorter part when divided by the centroid.

Notice that SW, QT and PO are the three medians of ΔSPQ, and V is the centroid
of ΔSPQ.

ΔSVQ is 1/3 of ΔSPQ.  Why? Remembering the formula for the area of a triangle,
A=1/2(bh), they have the same base SQ, and by theorem 2, the height OV of ΔSVQ
is 1/3 of the height OP of ΔSPQ.   

Since ΔSVQ is 1/3 of ΔSPQ which is 1/2 of the whole rhombus, then ΔSVQ is
(1/3)(1/2) = 1/6 of the whole rhombus.  

By theorem 2, VW is 1/2 of SV, and since ΔTVW is similar to ΔSVQ, the area
of ΔTVW, by theorem 1, is (1/2)2=1/4 the area of ΔSVQ
So ΔTVW is (1/4)(1/6) = 1/24th of the whole rhombus.

Therefore quadrilateral PWVT is 1/8 + 1/24 = 3/24 + 1/24 = 4/24 = 1/6 of rhombus PQRS.

Answer: 1/6
 
Edwin

Comment to greenestamps' solution.

While he is right that the problem does imply that the fraction is 
constant for all rhombuses, it is not obvious that it would be.  
I do think the student was expected to prove it is constant for all 
rhombuses, and not just find it for one particular case. 

You might suspect that I did draw the figure as the special case
greenestamps suggested. But I was careful to state that I could
not use that fact to prove it for all rhombuses. 
                         J

Edwin